Why Domain Conditions Must Be Combined
A formula involving several operations can place several restrictions on its inputs. A square root may require a radicand to be nonnegative, a logarithm requires a positive argument, and a denominator must not equal zero. When these operations appear together, satisfying just one restriction is not enough: an input is allowed only if every operation in the formula is defined.
The goal is to identify the inputs that meet all the conditions at once. Geometrically, each condition selects a region or set of points. The domain is the overlap of those sets, not their union.
The Restrictions to Combine
For the operations considered here, the basic conditions are as follows. A denominator must be nonzero. A square root of a real-valued expression is defined when its radicand is nonnegative. A logarithm of a real-valued expression is defined when its argument is strictly positive.
The placement of an operation can make its condition stricter. For example, \(\sqrt{q}\) in a numerator allows \(q=0\), but \(1/\sqrt{q}\) does not: the root must be defined and the resulting denominator must be nonzero. Thus \(q\) must be strictly positive when the square root is in a denominator.
A logarithm in a denominator has a related extra restriction. Its argument must be positive so that the logarithm exists, and the logarithm's value must be nonzero so that it can serve as a denominator. Since \(\ln t=0\) exactly when \(t=1\), a factor \(\ln(t)\) in a denominator requires \(t>0\) and \(t\ne1\).
Why Intersection Is the Right Rule
Suppose a formula has two parts, one defined on a set \(A\) of inputs and the other defined on a set \(B\). The whole formula can be evaluated only at an input where both parts can be evaluated. Those inputs are precisely the ones in \(A\cap B\), the intersection of the two sets. For more than two restrictions, the same reasoning applies repeatedly.
This gives a short justification of the method: evaluating a formula requires carrying out each operation it contains. If even one required operation is undefined, the formula has no real value at that input. Conversely, if every operation is defined and every denominator is nonzero, the stated operations can be carried out, so the formula is defined.
Worked Example: A Root, a Logarithm, and Two Denominator Factors
Worked Example: A Root, a Logarithm, and Two Denominator Factors
Find the domain of \(f(x,y)=\frac{\sqrt{6-x-y}}{(x-2)\ln(x+y-1)}\).
First, the square root requires
Next, the logarithm requires its argument to be positive:
There are two denominator factors. The factor \(x-2\) must be nonzero, so \(x\ne2\). The factor \(\ln(x+y-1)\) must also be nonzero. Because \(\ln t=0\) precisely when \(t=1\), this gives
All four conditions must hold. Therefore the domain consists of the points satisfying
Check an allowed point, \((x,y)=(3,0)\). The root radicand is \(6-3-0=3>0\); the logarithm argument is \(3+0-1=2>0\); the denominator factors are \(3-2=1\) and \(\ln2\ne0\). Substitution gives \(f(3,0)=\sqrt{3}/\ln2\), a defined real number.
Now check the boundary point \((0,6)\). The root radicand is \(6-0-6=0\), which is permitted. The logarithm argument is \(0+6-1=5>0\), and the denominator factors are \(-2\) and \(\ln5\), both nonzero. Thus this point is included, and \(f(0,6)=0/(-2\ln5)=0\). In contrast, at \((2,0)\) the log argument is \(1\), so the logarithm is zero and the denominator vanishes. That point is excluded.
Worked Example: A Logarithm and a Square Root in the Denominator
Worked Example: A Logarithm and a Square Root in the Denominator
Find the domain of \(g(x,y)=\frac{\ln(9-x^2-y^2)}{\sqrt{x-y+2}}\).
The logarithm argument must be strictly positive:
The square root is in the denominator. Its radicand must therefore be strictly positive: if it were zero, the denominator would be zero. Hence
The domain is the intersection of the disk interior \(x^2+y^2<9\) and the half-plane \(y<x+2\). Both strict inequalities are required.
At \((0,0)\), the logarithm argument is \(9-0-0=9>0\), and the denominator radicand is \(0-0+2=2>0\). Thus \(g(0,0)=\ln9/\sqrt2\) is defined. At \((3,0)\), the logarithm argument is \(9-9-0=0\), so the logarithm is undefined. At \((0,2)\), the logarithm argument is \(9-0-4=5>0\), but the denominator radicand is \(0-2+2=0\), so the quotient is undefined. Each test checks the original operations, not just one of the conditions.
Worked Example: Three Inputs and a Boundary That Is Allowed
Worked Example: Three Inputs and a Boundary That Is Allowed
Find the domain of \(h(x,y,z)=\frac{\sqrt{z-x-y}}{(z-2)\ln(x+y)}\).
The numerator's root requires \(z-x-y\ge0\), or \(z\ge x+y\). The logarithm requires \(x+y>0\). Since that logarithm is in the denominator, its value must also be nonzero; this excludes \(x+y=1\). The other denominator factor requires \(z\ne2\). Combining these conditions gives
Consider \((x,y,z)=(1,1,3)\). The root radicand is \(3-1-1=1\); the log argument is \(1+1=2>0\); \(z-2=1\ne0\); and \(\ln2\ne0\). Substitution gives \(h(1,1,3)=\sqrt1/(1\cdot\ln2)=1/\ln2\), which is defined.
The root boundary can be allowed: at \((1,1,2)\), the root radicand is \(2-1-1=0\), but \(z-2=0\), so the denominator vanishes and the point is excluded. At \((1,0,1)\), the root radicand is \(1-1-0=0\), and \(z-2=-1\ne0\), but the log argument is \(1\), making \(\ln1=0\). This point is also excluded. These checks show why a boundary permitted by one operation can still be ruled out by another.
A Common Mistake: Using “Or” Instead of “And”
Consider \(p(x,y)=\frac{\sqrt{x}}{(x-1)\ln y}\). The root requires \(x\ge0\), the factor \(x-1\) requires \(x\ne1\), and the logarithm in the denominator requires \(y>0\) and \(y\ne1\). The correct conditions are
A mistaken approach is to treat these as alternatives: for instance, to accept an input if either \(x\ge0\) or \(y>0\). That would accept \((-1,2)\) because \(2>0\). But substituting it into the original formula gives \(\sqrt{-1}\) in the numerator, which is not a real number. The input is not in the domain.
The correction is to test every condition. At \((-1,2)\), the condition \(x\ge0\) fails, so the point is excluded regardless of the fact that \(y>0\). In general, an intersection means “and”: each restriction must pass independently.
A Reliable Procedure
Identify every denominator, even root, and logarithm, including ones nested inside other expressions.
Use nonzero denominators, nonnegative radicands for roots outside denominators, and positive arguments for logarithms. Make a denominator root's radicand strictly positive.
Require all conditions simultaneously; do not replace “and” with “or.”
Test equality where a root allows zero, and inspect any factor that could make a denominator zero.
Substitute them into the original operations to confirm that each is defined.
Why the Combined Domain Matters
An incomplete domain can lead to invalid evaluations and misleading conclusions. In particular, boundary points deserve careful attention: a root may permit a zero radicand, while a denominator or logarithm elsewhere in the formula may exclude the same point. Conversely, a point on the boundary of a root condition should not be discarded automatically if every other operation remains defined.
The essential test is always the same: an input belongs to the domain exactly when every operation in the original formula has a valid real value there. Writing the separate restrictions first makes that test systematic and makes it easier to spot missing conditions.
Check Your Understanding
For each question, identify every restriction before combining them.
- What conditions must hold for a real square root that appears in a numerator?
- Why must the radicand of a square root in a denominator be strictly positive?
- For a factor \(\ln(q)\) in a denominator, what two conditions must hold for \(q\)?
- In the second worked example, why is \((0,2)\) excluded even though the logarithm is defined there?
- Find the domain conditions for \(r(x,y)=\frac{\sqrt{x+y}}{\ln(x-y)}\).